# Maximum degree of irreducible representation of group is less than or equal to product of maximum degree of irreducible representation of subgroup and index of subgroup

## Statement

Suppose $G$ is a finite group, $H$ is a subgroup, and $K$ is a field whose characteristic does not divide the order of $G$ (we do not require $K$ to be a splitting field, though the splitting field case is of particular interest).

Then, the Maximum degree of irreducible representation (?) of $G$ over $K$ is less than or equal to the product:

(maximum degree of irreducible representation of $H$ over $K$) times (index of $H$ in $G$, i.e., $[G:H]$)

## Proof

### Proof in characteristic zero

Given: $G$ is a finite group, $H$ is a subgroup, and $K$ is a field of characteristic zero. $H$ has index $d$ in $G$. $m$ is the maximum of the degrees of irreducible representations of $G$ over $K$.

To prove: $H$ has an irreducible representation over $K$ whose degree is at least $m/d$.

Proof:

Step no. Assertion/construction Facts used Given data used Previous steps used Explanation
1 Let $\alpha$ be the character of an irreducible representation of $G$ of degree $m$. $m$ is the maximum of degrees of irreducible representations of $G$.
2 Let $\beta$ be the character of an irreducible representation of $H$ arising as a subrepresentation of $\operatorname{Res}_H^G \alpha$. $H$ is a subgroup of the finite group $G$.
3 The inner product $\langle \operatorname{Res}_H^G \alpha, \beta \rangle_H$ is a positive integer, because $\beta$ is the character of a subrepresentation of the representation affording $\operatorname{Res}_H^G\alpha$. (Note: Over a splitting field, the inner product is precisely the multiplicity, but it may be bigger in general if $\beta$ is not absolutely irreducible). We need $K$ to have characteristic zero to make sense of positive integer.
4 $\langle \operatorname{Res}_H^G \alpha, \beta \rangle_H = \langle \alpha, \operatorname{Ind}_H^G \beta \rangle$ Fact (2)
5 $\langle \alpha, \operatorname{Ind}_H^G \beta \rangle$ is a positive integer, indicating that $\alpha$ is an irreducible constituent of $\operatorname{Ind}_H^G \beta$. Steps (3), (4)
6 The degree of $\alpha$, which is $m$, is less than or equal to the degree of $\operatorname{Ind}_H^G\beta$ Steps (1), (5)
7 The degree of $\operatorname{Ind}_H^G\beta$ is $d$ times the degree of $beta$ $H$ has index $d$ in $G$. Definition of induced representation.
8 The degree of $beta$ is at least $m/d$ Steps (6), (7)